Alternative Group Theory Approach

I am trying to learn basic group theory and I have a basic question.

Groups are defined in terms of a binary operator and and elements along with their inverses. In category theory, Grp denotes this kind of structures. Generally, the binary operation is commutative like + or *. Inverses are defined in terms of this operator and they closely related. (-a or 1/a)

My question is, why elements are defined as a/-a, not just a's? Let me explain. In computation theory perspective, searching fewer elements is always advantageous. Once group elements are defined as a and -a, a search algorithm has to make to pass all elements twice.

Ok, let me build another structure that is very similar but different.

Let (-) be a binary the operator which is not commutative. It is obvious that two minuses would be a + so, f^2 is (+) a different operator.

So, defining a different operator, no need to define inverses in any sense. It is the operator that makes the difference.

This might sound awkward but I see groups as elements and an operator whose order is 2.

Ring are some elements and two operators, one has order 2, the other is 1. Fields are some elements with two operators both having order 2.

I am not trying to convince Math people that this is more finer, I am just looking for alternative theories that defines similar structures of Group Theory.

Any references would be much appreciated.

Edit #1:

$a - b = a - b - (0 - 0)$

$a + b = a - 0 - (0 - b)$

In matrix notation, they may be denoted as:

$\left( \begin{array}{ccc} a & b \\ 0 & 0 \end{array} \right)$

and

$\left( \begin{array}{ccc} a & 0 \\ 0 & b \end{array} \right)$

Defined operators above can be used to construct two element set. Yet building a group structure from those two is a question still standing.

Regards.

-
I don't understand the question. Most groups are not commutative, and I don't see what the existence of inverses has to do with passing through all the elements twice. –  Qiaochu Yuan Mar 5 '12 at 22:00
What is $f^2$? If $(-)$ is a binary operator, then how do I take two of it? –  Dylan Moreland Mar 5 '12 at 22:17
The question, "why are elements defined as $a/-a$?" assumes facts not in evidence. Cite, please, with exact quotation, a source where you have seen an element defined as "$a/-a$". I also can't make much sense out of the only other thing that is requested, "alternative theories that defines (sic) similar structures of Group Theory." There might be a serious question here, but it's going to take a lot of thought and rewriting to excavate it. –  Gerry Myerson Mar 5 '12 at 22:46

Suppose we have a finite group G. What does that mean? It means we have a binary map $m: G\times G \to G$ that's associative, an identity $1$ with respect to the $m$ map, and a map $i:G \to G$ with the property that $i(g) g = 1$ for all $g \in G$.
You might note that no elements were defined as "a's/-a's$, but instead that we simply have a set, and we know that elements correspond in (possibly trivial) pairs. So if I were to search through a group, I would not see any reason to check every element, and then check its inverse. In fact, checking every element is the same as checking the inverse of every element. If we have a binary operator$f:S\times S \to S$, then we might have for instance that$f(a,b) = c$. With groups, we might represent this by creating the notation$m(g,h) =: g \cdot h =: gh = j$(multiplicatively) or$g + h$(additively) with some starting$g,h \in G$and some resulting$j \in G$. As such, the notation$f^2$doesn't carry any meaning. It's not literally$ f \circ f$, because then we would try to do something like$f \circ f (a,b) = f(c,?)$where, as before, we suppose$f(a,b) = c$. There will never be an 'order 2 binary operator' in that sense, because a binary operator takes in two arguments and only returns 1. On the other hand, we look at the 'negative' notation. I admit, this looks confusing. Addition is a binary operator. Why does$-(-1)$make sense? Well, the problem is that the$-$in$-a$corresponds to inversion, which is a unitary operator (I called it$i$above). The great fluke is that$a - b$in a group actually means$a + (-b)$. And inversion is of order 2 in the sense that$i(i(x)) = x$for all$x \in G$. As for a different formulation - did you know that an associative binary operator$m$, a left identity$e$s.t.$m(e,g) = g$for all$g$, and a left-inversion map$i$s.t.$i(g) g = e$for all$g$is all that's required for the resulting set to be a group? But changing it just a little bit, such as having a right identity$e$s.t.$m(g,e) = g$for all$g$but a left-inversion map (as above) does not make a group. These are very finicky rules sometimes. In addition, there are simpler things. A set with a binary operation, with or without an identity, closed, with or without inversion... these motivate the ideas behind monoids and semigroups, which are two weaker sets than groups. Similarly, stronger sets include rings, fields, modules, algebras and ring-algebras, and vector spaces. (And probably more, but these are the things that come to mind). And since you mentioned the word category, I should also mention so-called Group Objects, which are strange things. I'm still a bit afraid of category theory, so I don't want to try to say much more about that. I would also like to note that many useful groups are infinite in size (almost all that I use with any reasonable frequency), and brute-force search algorithms on the groups are not a good approach to general group theoretic search algorithms. - Thanks mixedmath, I appreciate your answer. The paragraph, As for a different formulation - did you know that an associa.. is very similar to my ambition. For instance, 0 is only a right identity in case of (-). Yet I haven't seen a right-identity definition taken seriosly in any algebra book. – foobar Mar 5 '12 at 23:32 @foobar: You haven't looked at the right textbooks. The kind of structure that contemplates a binary nonassociative operator is called a "magma"; in semigroups, loops, and magmas, discussion of one-sided identities is common. – Arturo Magidin Mar 6 '12 at 4:05 @mixedmath I tried to answer your$f^{2}\$ question by defining operators in my mind just using subtraction. Division can be defined in a very similar fashion. –  foobar Mar 6 '12 at 19:17